题解 | #获得积分最多的人(三)#

select
    id,
    name,
    grade_num
from (
        select
        user_id,
        grade_num,
        rank() over (order by grade_num desc) as r1
        from (
            select
                user_id,
                sum(if(type = 'add',grade_num,-1 * grade_num)) as grade_num
            from grade_info
            group by user_id
        ) as g1
    ) as g2
join user u
on u.id = g2.user_id
where g2.r1 = 1
group by id,name